The probability distribution of the random variable X is given by
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | 0.2 | k | 2k | 2k |
Find the variance of the random variable \(X\).
Step 1: Conceptual Understanding:
This question assesses the ability to calculate the variance of a discrete random variable from its probability distribution. The process involves first determining the unknown constant 'k' by ensuring the sum of all probabilities equals 1. Subsequently, the mean ( \( E[X] \) ) and the expected value of the squared variable ( \( E[X^2] \) ) are computed. The variance is then derived using the standard formula.
Step 2: Essential Formulas and Methodology:
1. Probability Summation: \( \sum P(X_i) = 1 \).
2. Mean (Expected Value): \( E[X] = \mu = \sum X_i P(X_i) \).
3. Expected Value of \( X^2 \): \( E[X^2] = \sum X_i^2 P(X_i) \).
4. Variance Formula: \( \text{Var}(X) = \sigma^2 = E[X^2] - (E[X])^2 \).
Step 3: Detailed Calculation:
Determination of k:
The sum of probabilities must equal 1. \[ 0.2 + k + 2k + 2k = 1 \] \[ 0.2 + 5k = 1 \] \[ 5k = 0.8 \implies k = \frac{0.8}{5} = 0.16 \] The completed probability distribution table is as follows:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | 0.2 | 0.16 | 0.32 | 0.32 |
Calculation of Mean \( E[X] \):
\[ E[X] = (0 \times 0.2) + (1 \times 0.16) + (2 \times 0.32) + (3 \times 0.32) \] \[ E[X] = 0 + 0.16 + 0.64 + 0.96 = 1.76 \]
Calculation of \( E[X^2] \):
\[ E[X^2] = (0^2 \times 0.2) + (1^2 \times 0.16) + (2^2 \times 0.32) + (3^2 \times 0.32) \] \[ E[X^2] = (0 \times 0.2) + (1 \times 0.16) + (4 \times 0.32) + (9 \times 0.32) \] \[ E[X^2] = 0 + 0.16 + 1.28 + 2.88 = 4.32 \]
Calculation of Variance \( \text{Var}(X) \):
\[ \text{Var}(X) = E[X^2] - (E[X])^2 \] \[ \text{Var}(X) = 4.32 - (1.76)^2 \] \[ \text{Var}(X) = 4.32 - 3.0976 = 1.2224 \]
Conversion to Fraction:
Comparing with the provided options.
Option (A): \( \frac{764}{625} \).
The decimal value of this fraction is: \( 764 \div 625 = 1.2224 \).
This value matches the calculated variance.
Step 4: Final Result:
The variance of the random variable \( X \) is 1.2224, which is equivalent to \( \frac{764}{625} \).
If the mean and the variance of the data 
are $\mu$ and 19 respectively, then the value of $\lambda + \mu$ is
In the figure, a sector of the circle with central angle 120° is given. If a dot is put in the circle without looking, what is the probability that the dot is in the shaded region ?